When a sequence $f \in \ell_1$, there is a very simple bound on its $\ell_q$-norms given by $\|f\|_q^q \leq \|f\|_1 \cdot \|f\|_\infty^{q-1}$.
This inequality is a special (or rather limit) case of Lyapounov's inequality: let $1 \leq p,q < \infty$ and $0 \leq \lambda \leq 1$. If $r = \lambda p + (1 - \lambda)q$ and $f \in \ell_p \cap \ell_q $, then
$$||f||_r^r \leq ||f||_p^{\lambda p} ||f||_q^{(1 - \lambda)q}$$
I am looking for the corresponding inequality for Orlicz norms:
Question: Given a Young functionfunctions $\phi$$\phi_1, \phi_2$ and $\phi_3$, and $f \in \ell_1$$f \in \cap_i \ell_{\phi_i}$ is there a bound of the form $\|f\|_\phi \leq $$\|f\|_{\phi_2} \leq $ some expression with $\|f\|_1$$\|f\|_{\phi_1}$ and $\|f\|_\infty$$\|f\|_{\phi_3}$ (so that the "expression" tends to 0 if $\|f\|_\infty$one of the norms tends to 0 and $\|f\|_1$while the other remains bounded]bounded).
Naively, one would bound:The limit case $\sum \phi(|x|) \leq \sum |x| \frac{\phi(|x|)}{|x|} \leq \big( \sup_x \frac{\phi(|x|)}{|x|} \big) \sum |x|$. But I am quite sure this is neither optimal nor correct(where $\phi_1$ and $\phi_3$ are the 1 and $\infty$ norms) could be done as follows (given thatunder the actual normassumption that $\frac{\phi(x)}{x}$ is defined duallyincreasing and does not take the value $+\infty$).
Note that It's easier to do it with the inequality I mentionedLuxemburg norm (which is the same as the original one up to a special casefactor of Lyapounov's inequality: let2). Let $1 \leq p,q < \infty$ and$f \in \ell_1 \cap \ell_\infty$, then $$ \| f\|_\phi = \inf \lbrace b >0 \mid \sum_n \phi( \tfrac{|f(n)|}{b}) \leq 1 \rbrace $$ Bound: $0 \leq \lambda \leq 1$$\sum \phi(|x|) \leq \sum |x| \tfrac{\phi(|x|)}{|x|} \leq \big( \sup_x \frac{\phi(|x|)}{|x|} \big) \sum |x|$. If $r = \lambda p + (1 - \lambda)q$ andAssuming that $f \in \ell_p \cap \ell_q $$\phi(x)/x$ is increasing, thenthis means that
$$||f||_r^r \leq ||f||_p^{\lambda p} ||f||_q^{(1 - \lambda)q}$$$$ \sum_n \phi( \tfrac{|f(n)|}{b}) \leq \frac{\phi(\|f\|_\infty / b)}{\|f\|_\infty} \|f\|_1. $$
SoIn particular, if $b = \dfrac{\|f\|_\infty}{ \phi^{-1} \big(\tfrac{\|f\|_\infty}{\|f\|_1} \big) }$, where $\phi^{-1}$ is a reference to this larger context could also be niceinverse (how to bound one Orlicz norm from two othersor semi-inverse if $\phi$ is not strictly increasing) of $\phi$, the sum is smaller than one. Hence $$ \|f\|_\phi \leq \dfrac{\|f\|_\infty}{ \phi^{-1} \big(\tfrac{\|f\|_\infty}{\|f\|_1} \big) } $$
EDIT: someSome further comments:
If it helps remove some technicalities, I would be fine with thevarious extra assumption thatassumptions such as $\phi: [0,\infty[ \to [0,\infty[$ is$\phi_i: [0,\infty[ \to [0,\infty[$ are strictly increasing.
The above inequalities are also true in the $L_p$-spaces (if you add the obvious assumptions like $f \in L_1 \cap L_\infty$). They can be proved using Hölder's inequality (although $\|f\|_q^q \leq \|f\|_1 \|f\|_\infty^{q-1}$ can be proved using a much more naive argument [similar to the one above])
I found Hölder inequalities for Orlicz norms (as well as some inequalities about convolutions), but I could not find the above "interpolation" inequalities.